- \frac{u\epsilon}{\nu}\sim 1.
Assuming the smallest scale carries a substantial part of the total dissipation gives
- \nu(\frac{u}{\epsilon})^2\sim 1.
Combination gives
- u\sim \nu^{\frac{1}{4}}
- \epsilon\sim\nu^{\frac{3}{4}}
suggesting that the turbulent solution is Lipschitz continuous with exponent \frac{1}{3}.
My question to Tao posed as a post comment is if according to the Clay problem formulation, such a Lip^\frac{1}{3} turbulent solution with smallest scale \nu^\frac{3}{4} is to be viewed as a smooth solution for any small \nu >0?
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